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yKpoI14uk [10]
3 years ago
5

HELP ASAP!!! 2C4H10(g) + 1302 ===> 8CO2(g) + 10H2O(g)

Chemistry
1 answer:
Ivan3 years ago
6 0

Answer:

30.34g (corrected to 4 significant figures).

Explanation:

Take the atomic mass of C=12.0, H=1.0, O=16.0.

no. of moles = mass / molar mass

So, no. of moles of butane reacted = 10 / (12x4 + 1x10)

= 0.172414 mol

Since O2 is in excess and butane is the limiting reagent, the no. of moles of carbon dioxide produced depends on the no. of moles of butane reacted.

From the equation, the mole ratio of butane:Carbon dioxide = 2: 8 = 1: 4,

meaning 1 mole of butane gives 4 moles of CO2.

Using this ratio,

we can deduce that the no. of moles of CO2 produced =  0.172414 x 4

=0.689655 mol

As mass = no. of moles x molar mass

mass of CO2 produced =   0.689655 x (12.0+16.0x2)

=30.34g (corrected to 4 significant figures).

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Daniel [21]

Answer:

Percentage yield = 30%

Explanation:

Given data:

Number of moles of NO = 7.0 mol

Number of moles of O₂ = 5 mol

Number of moles of NO₂ = 3 mol

Percentage yield = ?

Solution:

Chemical equation:

2NO + O₂ → 2NO₂

Now we will compare the moles of NO₂ with NO and O₂ .

                  NO           :               NO₂

                  2               :               2

                 7.0             :              7.0

                O₂               :                NO₂

                 1                 :                 2

                 5.0             :               2 ×5.0 = 10 mol

The number of moles of NO₂ produced by NO are less it will be limiting reactant.

Mass of NO₂ = moles × molar mass

Mass of NO₂ = 10 mol × 46g/mol

Mass of NO₂ =  460 g

Actual yield of NO₂:

Mass of NO₂ = moles × molar mass

Mass of NO₂ = 3 mol × 46g/mol

Mass of NO₂ =  138 g

Percentage yield:

Percentage yield = Actual yield/theoretical yield × 100

Percentage yield = 138 g/ 460 g × 100

Percentage yield = 30%

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Answer:

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Explanation:

Given data:

[OH⁻] = 9.50 ×10⁻⁹M

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